Soluţii corecte: Ady Nicolae, Emil Claudiu Man, Aurel Ionescu, Ionel-Vasile Pit-Rada, Zoltan Szabo, Marius Stratulat, Viorel Manta, Cristian Daniel Balanoiu

Problema admite solutii cu un numar foarte mic de cifre 3 in intreaga expresie matematica, chiar si 1 singura cifra de 3. Exemplificam mai jos cateva din solutiile primite, printre care unele cu aceasta proprietate.

 

Ionel-Vasile Pit-Rada:

O expresie care foloseste 7 cifre egale cu 3 se poate obtine astfel:
2018=2160-142
2160=3!!*3

2=[sqrt(3!)]
5=[sqrt(sqrt((3!)!))]
29=24+5=4!+5=([sqrt(3!)]+[sqrt(3!)])!+[sqrt(sqrt((3!)!))]
142=145-3=5*29-3=[sqrt(sqrt((3!)!))]*(([sqrt(3!)]+[sqrt(3!)])!+[sqrt(sqrt((3!)!))])-3
Astfel obtinem
2018=(3!)!*3-[sqrt(sqrt((3!)!))]*(([sqrt(3!)]+[sqrt(3!)])!+[sqrt(sqrt((3!)!))])+3

Pentru o expresie care sa foloseasca 6 cifre egale cu 3 putem proceda astfel:
2018=45^2-7
45=3*3*[sqrt(sqrt((3!)!))]
45^2=(3*3*[sqrt(sqrt((3!)!))])^[sqrt(3!)]
7=5+2=[sqrt(sqrt((3!)!))]+[sqrt(3!)]
2018=(3*3*[sqrt(sqrt((3!)!))])^[sqrt(3!)]-[sqrt(sqrt((3!)!))]-[sqrt(3!)]

O solutie care foloseste doar 5 cifre de 3:

2018=([sqrt([sqrt(sqrt((3!)!))]!)]^3+3*3)*[sqrt(3!)]

O expresie cu o singura cifra 3, pentru calcularea lui 2018.

Facem urmatoarea notatie:  f(a,s)=partea intreaga din radacina patrata aplicata de s ori pentru factorial din a

2018=f(f(f(f(f(f(f(f(f(f(f(f(f(f(3,0),1),4),5),6),4),2),22),9),8),15),19),21),12)

Avem astfel o expresie din care se obtine 2018 folosind o singura data cifra 3.

In expresie se aplica functia parte intreaga de 13 ori, functia factorial de 14 ori si functia sqrt() de 128 de ori

Explicatie:
f(3,0)=6
f(6,1)=26
f(26,4)=46
f(46,5)=63
f(63,6)=23
f(23,4)=25
f(25,2)=1984547
f(1984547,22)=594
f(594,9)=521
f(521,8)=44894
f(44894,15)=600944
f(600944,19)=1336501
f(1336501,21)=4239
f(4239,12)=2018

Verificarea se poate face cu un program C/C++ sau in online pas cu pas in http://www.wolframalpha.com/

 

Zoltan Szabo:

O solutie ce foloseste numai 3 cifre de 3 si poate fi verificata in Excel:

[sqrt([sqrt([sqrt(sqrt((3!)!))]!)]!)]+[sqrt(sqrt((3!)!))]!-3!

INT(SQRT(FACT(INT(SQRT(FACT(INT(SQRT(SQRT(FACT(FACT(3)))))))))))+FACT(INT(SQRT(SQRT(FACT(FACT(3))))))-FACT(3)

 

Am gasit 2 solutii cu 2 cifre cu un program C++ bazat pe algoritmul lui Lee.

Solutia 1:

Am folosit urmatoarele notatii
fact(n) = n factorial
sqrtx(n) = partea intreaga a extragerii radicalului de ordin 2 din n efectuat de x ori 
Solutia 1:
2018=29+1989
Fiecare din cele doua numere se poate scrie cu o singura cifra de 3.
fact(3)=6
fact(6)=720
sqrt9(fact(720))=2576
sqrt13(fact(2576))=8
sqrt(fact(8))=200
sqrt8(fact(200))=29

sqrt8(fact(sqrt(fact(sqrt13(fact(sqrt9(fact(fact(fact(3))))))))))=29

fact(3)=6
sqrt2(fact(6))=5
sqrt(fact(5))=10
sqrt2(fact(10))=43
sqrt4(fact(43))=1989

sqrt4(fact(sqrt2(fact(sqrt(fact(sqrt2(fact(fact(3)))))))))=1989

Deci avem
2018=sqrt8(fact(sqrt(fact(sqrt13(fact(sqrt9(fact(fact(fact(3))))))))))+
     sqrt4(fact(sqrt2(fact(sqrt(fact(sqrt2(fact(fact(3)))))))))

Solutia 2:
2018=2027-9

fact(3)=6
fact(6)=720
sqrt9(fact(720))=2576
sqrt11(fact(2576))=5560
sqrt12(fact(5560))=31237
sqrt17(fact(31237))=9

sqrt17(sqrt12(fact(sqrt11(fact(sqrt9(fact(fact(fact(3)))))))))=9

fact(3)=6
fact(6)=720
sqrt10(fact(720))=50
sqrt4(fact(50))=10719
sqrt13(fact(10719))=50742
sqrt16(fact(50742))=2027

sqrt16(fact(sqrt13(fact(sqrt4(fact(sqrt10(fact(fact(fact(3))))))))))=2027

Deci avem
2018=sqrt16(fact(sqrt13(fact(sqrt4(fact(sqrt10(fact(fact(fact(3))))))))))-
     sqrt17(sqrt12(fact(sqrt11(fact(sqrt9(fact(fact(fact(3)))))))))

 

Solutie cu 1 singura cifra:

Am folosit urmatoarele notatii
fact(n) = n factorial
sqrtx(n) = partea intreaga a extragerii radicalului de ordin 2 din n efectuat de x ori 
sqrt1(n) am notat cu sqrt(n)

fact(3)=6
sqrt(fact(6))=26
sqrt4(fact(26))=46
sqrt5(fact(46))=63
sqrt6(fact(63))=23
sqrt4(fact(23))=25
sqrt2(fact(25))=1984547
sqrt22(fact(1984547))=594
sqrt9(fact(594))=521
sqrt8(fact(521))=44894
sqrt15(fact(44894))=600944
sqrt19(fact(600944))=1336501
sqrt21(fact(1336501))=4239
sqrt12(fact(4239))=2018

sqrt12(fact(sqrt21(fact(sqrt19(fact(sqrt15(fact(sqrt8(fact(sqrt9(fact(sqrt22(fact(sqrt2(fact(sqrt4(fact(sqrt6(fact(sqrt5(fact(sqrt4(fact(sqrt(fact(fact(3)))))))))))))))))))))))))))=2018

 

O alta solutie cu 1 singura cifra:

Am folosit urmatoarele notatii
fact(n) = n factorial
sqrtx(n) = partea intreaga a extragerii radicalului de ordin 2 din n efectuat de x ori 
sqrt1(n) am notat cu sqrt(n)

fact(3)=6
fact(6)=720
sqrt10(fact(720))=50
sqrt4(fact(50))=10719
sqrt15(fact(10719))=15
sqrt3(fact(15))=32
sqrt4(fact(32))=163
sqrt6(fact(163))=35611
sqrt15(fact(35611))=29825
sqrt15(fact(29825))=4758
sqrt12(fact(4758))=5859
sqrt12(fact(5859))=58735
sqrt16(fact(58735))=7670
sqrt12(fact(7670))=2896254
sqrt22(fact(2896254))=14525
sqrt14(fact(14525))=2018

sqrt14(fact(sqrt22(fact(sqrt12(fact(sqrt16(fact(sqrt12(fact(sqrt12(fact(sqrt15(fact(sqrt6(fact(sqrt4(fact(sqrt3(fact(sqrt15(fact(sqrt4(fact(sqrt10(fact(fact(fact(3))))))))))))))))))))))))))))=2018

 

Extindere: Ce sa intampla daca in loc de cifra 3 problema cerea folosirea unei alte cifre?

Zoltan Szabo:

Solutii cu diferite cifre:
Dupa ce am gasit o solutie cu cifra 3, destul de usor se poate trece de la o cifra la alta. Trebuie sa gasim trecerea de la cifra c la cifra 3 si sa inlocuim aceasta valoare, sau in lantul de apeluri de radicali-factorial sa regasim cifra c (de exemplu fact(3)=6) .

Cu cifrele 1 si 2 nu exista solutie cu o cifra, factorialul este punct fix, iar sirul radicalilor converge descrescator catre 1.

Cu cifra 4:
Avand in vedere ca
fact(4)=24
sqrt5(fact(24))=5
sqrt2(fact(5))=3
avem sqrt2(fact(sqrt5(fact(fact(4)))))=3

Cu cifra 5:

Avem sqrt2(fact(5))=3

Cu cifra 6:

Avem fact(3)=6

Cu cifra 7:

sqrt2(fact(7))=8
sqrt3(fact(8))=3
Avem sqrt3(fact(sqrt2(fact(7))))=3

Cu cifra 8:

Avem sqrt3(fact(8))=3

Cu cifra 9:

Avem sqrt(9)=3